$\int_{0}^{\pi} x f(sin x) dx $ is equal to

Q: $\displaystyle \int_{0}^{\pi} x f(sin x) dx $ is equal to

(A) $ \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

(B) $ \pi \int_{0}^{\pi/2} f(sin x) dx $

(C) $ 2 \pi \int_{0}^{\pi/2} f(sin x) dx $

(D) none of these

Ans: (A) , (B)

Sol: Let $\displaystyle I = \int_{0}^{\pi} x f(sin x) dx $

$\displaystyle = \int_{0}^{\pi} (\pi – x ) f sin (\pi-x) dx $

$\displaystyle I = \pi \int_{0}^{\pi} f(sin x) dx – I $

$\displaystyle 2 I = \pi \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = \frac{\pi}{2} \int_{0}^{\pi} f(sin x) dx $

$\displaystyle I = 2 \times \frac{\pi}{2} \int_{0}^{\pi/2} f(sin x) dx $

$\displaystyle I = \pi \int_{0}^{\pi/2} f(sin x) dx $

Hence (A) and (B) are the correct answers.